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Java Program For Selecting A Random Node From A Singly Linked List

Given a singly linked list, select a random node from the linked list (the probability of picking a node should be 1/N if there are N nodes in the list). You are given a random number generator.
Below is a Simple Solution:

1. Count the number of nodes by traversing the list.
2. Traverse the list again and select every node with probability 1/N. The selection can be done by generating a random number from 0 to N-i for i’th node, and selecting the i’th node only if the generated number is equal to 0 (or any other fixed number from 0 to N-i).

We get uniform probabilities with the above schemes.

```i = 1, probability of selecting first node = 1/N
i = 2, probability of selecting second node =
[probability that first node is not selected] *
[probability that second node is selected]
= ((N-1)/N)* 1/(N-1)
= 1/N  ```

Similarly, probabilities of other selecting other nodes is 1/N
The above solution requires two traversals of linked list.

How to select a random node with only one traversal allowed?
The idea is to use Reservoir Sampling. Following are the steps. This is a simpler version of Reservoir Sampling as we need to select only one key instead of k keys.

```(1) Initialize result as first node
(2) Initialize n = 2
(3) Now one by one consider all nodes from 2nd node onward.
(a) Generate a random number from 0 to n-1.
Let the generated random number is j.
(b) If j is equal to 0 (we could choose other fixed numbers
between 0 to n-1), then replace result with the current node.
(c) n = n+1
(d) current = current->next```

Below is the implementation of above algorithm.

Java

 `// Java program to select a random``// node from singly linked list``import` `java.util.*;` `// Linked List Class``class` `LinkedList``{``    ``// head of list``    ``static` `Node head; ` `    ``// Node Class``    ``static` `class` `Node``    ``{``        ``int` `data;``        ``Node next;` `        ``// Constructor to create``        ``// a new node``        ``Node(``int` `d)``       ``{``            ``data = d;``            ``next = ``null``;``        ``}``    ``}` `    ``// A reservoir sampling-based function``    ``// to print a random node from a``    ``// linked list``    ``void` `printrandom(Node node)``    ``{``        ``// If list is empty``        ``if` `(node == ``null``)``        ``{``            ``return``;``        ``}` `        ``// Use a different seed value so``        ``// that we don't get same result``        ``// each time we run this program``        ``Math.abs(UUID.randomUUID().getMostSignificantBits());` `        ``// Initialize result as first node``        ``int` `result = node.data;` `        ``// Iterate from the (k+1)th element``        ``// to nth element``        ``Node current = node;``        ``int` `n;``        ``for` `(n = ``2``; current != ``null``; n++)``        ``{``            ``// change result with``            ``// probability 1/n``            ``if` `(Math.random() % n == ``0``)``            ``{``                ``result = current.data;``            ``}` `            ``// Move to next node``            ``current = current.next;``        ``}``        ``System.out.println(``        ``"Randomly selected key is "` `+ result);``    ``}` `    ``// Driver code``    ``public` `static` `void` `main(String[] args)``    ``{``        ``LinkedList list = ``new` `LinkedList();``        ``list.head = ``new` `Node(``5``);``        ``list.head.next = ``new` `Node(``20``);``        ``list.head.next.next = ``new` `Node(``4``);``        ``list.head.next.next.next = ``new` `Node(``3``);``        ``list.head.next.next.next.next = ``new` `Node(``30``);``        ``list.printrandom(head);``    ``}``}``// This code is contributed by Mayank Jaiswal`

Time Complexity: O(n), as we are using a loop to traverse n times. Where n is the number of nodes in the linked list.
Auxiliary Space: O(1), as we are not using any extra space.

Note that the above program is based on the outcome of a random function and may produce different output.

How does this work?
Let there be total N nodes in list. It is easier to understand from the last node.
The probability that the last node is result simply 1/N [For last or N’th node, we generate a random number between 0 to N-1 and make the last node as a result if the generated number is 0 (or any other fixed number]
The probability that second last node is result should also be 1/N.

```The probability that the second last node is result
= [Probability that the second last node replaces result] X
[Probability that the last node doesn't replace the result]
= [1 / (N-1)] * [(N-1)/N]
= 1/N```

Similarly, we can show probability for 3rd last node and other nodes. Please refer complete article on Select a Random Node from a Singly Linked List for more details!